Question: Simplify the following expression and state the condition under which the simplification is valid. $x = \dfrac{-8q^2 - 8q + 16}{q^3 + 5q^2 - 6q}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ x = \dfrac {-8(q^2 + q - 2)} {q(q^2 + 5q - 6)} $ $ x = -\dfrac{8}{q} \cdot \dfrac{q^2 + q - 2}{q^2 + 5q - 6} $ Next factor the numerator and denominator. $ x = - \dfrac{8}{q} \cdot \dfrac{(q - 1)(q + 2)}{(q - 1)(q + 6)}$ Assuming $q \neq 1$ , we can cancel the $q - 1$ $ x = - \dfrac{8}{q} \cdot \dfrac{q + 2}{q + 6}$ Therefore: $ x = \dfrac{ -8(q + 2)}{ q(q + 6)}$, $q \neq 1$